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\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
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\begin{document}
\date{}
\section*{2014年数学分析}
\begin{problem}[本题30分,每题10分]\\
    1)求极限$\displaystyle\lim_{x\to0}{\frac{1}{\sin^{4}x}}\int_{0}^{x^{2}}\tan t\ d t$\\
    2)求积分$\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{1+\sin^{2}\theta}\d\theta $\\
    3)计算曲线积分$\int_{L}\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}}\ d y$,其中$L$为曲线$y=x^{2}-1$自$A(-1,0)$至$B(2,3)$的弧段.\\
\end{problem}
\begin{solution}[本题30分,每题10分]\\
    1)$\displaystyle\lim_{x\to0}\frac{\int_{0}^{x^2}\tan t \ d t}{\sin^{4}x}=\displaystyle\lim_{x\to0}\frac{2x\tan x^{2}}{4\sin^{3}x\cdot\cos x}\\
    =\displaystyle\lim_{x\to0}\frac{\tan x^{2}}{2\sin^{3}x\cos x}=\displaystyle\lim_{x\to0}\frac{\sec^{2}x^{2}\cdot2x}{4\sin x\cos^{3}x-2\sin^{3}x}\\
    =\displaystyle\lim_{x\to0}\frac{\sec^{2}x^{2}}{2(\cos^{2}x-\sin^{2}x)}=\frac{1}{2}$\\
    2)$\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{1+\sin^{2}\theta}\ d \theta \\
    =\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin^2\theta}\ d \sin\theta \\
    =\arctan \sin\theta\bigg|_{0}^{\frac{\pi}{2}}=\frac{\pi}{4}$\\
    3)$f_{L}\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}} \ d y\\
    =\oint\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}}\ d y-\int_{(1,0),(2,3)}\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}}\ d y\\
    =-\int_{m}\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}}\ d y $\\
    直线的参数方程为$\begin{cases}x=-1+3t\\y=3t\end{cases}$,$t \in[0,1]$\\
    $\therefore
    \begin{aligned}
        &\int_{m}\frac{-y}{x^{2}+y^{2}}\ d x+\frac{x}{x^{2}+y^{2}}\ d y \\
        &=\int_{0}^{1}\left[\frac{-3t\cdot3}{(3t-1)^{2}+(3t)^{2}}+\frac{3t-1}{(3t-1)^{2}+(3t)^{2}}\cdot3\right]\ d t \\
        &=\int_{0}^{1}\frac{-3}{18t^{2}-6t+1}\ d t \\
        &=-3\int_{0}^{\prime}\frac{1}{(3\sqrt{2}t-\frac{1}{\sqrt{2}})^{2}+\frac{1}{2}}\ d t\\
        &=-6\int_{0}^{1}\frac{1}{(bt-1)^{2}+1}\ d t \\
        &=-\int_{0}^{\prime}\frac{1}{(6t-1)^{2}+1}\ d (6t-1) \\
        &=-\arctan6t-1\int_{0}^{1}=-\arctan5-\frac{\pi}{4}
    \end{aligned}
    $
\end{solution}


\begin{problem}[本题10分]
    设函数$f(x)$在$(0,+\infty)$上有定义,$f(x^{2})=f(x)$,且$\displaystyle\lim_{x\to0^{+}}f(x)=\displaystyle\lim_{x\to+\infty}f(x)=f(1)$,
证明:$f(x)$为常量函数，$x\in (0,+\infty)$.\\
\end{problem}
\begin{solution}[本题10分]\\
    证明：\\
    $\because$ $f(x^{2})=f(x)$\\
    $\therefore$ $f(x^{m})=f(x)$\\
    当$x=1$时，$f(x)=f(1)$\\
    当$x>1$时，$f(x)=f(m)=\displaystyle \lim_{x\to+\infty}f(x)=f(1)$\\
    当$x<1$时，$f(x)=f(m)=\displaystyle \lim_{x\to0^{+}}f(x)=f(1)$\\
    $\therefore$ $f(x)\equiv f(1)$\\
    $\therefore$ $f(x)$为常量函数
\end{solution}


\begin{problem}[本题10分]
    设函数$f(x)$在$[0,+\infty)$上可微,且$0\leq f^{\prime}(x)\leq f(x)$,$f(0)=0$,
证明：在$[0,+\infty)$上$f(x)\equiv0$.
\end{problem}
\begin{solution}[本题10分]\\
证明:\\
$\because$ $f(x)$在$[0,+\infty)$上可微\\
$\therefore$ $f(x)$在$[0,\frac{1}{2})$上可微\\
$\therefore$ $f(x)$在$[0,\frac{1}{2})$连续\\
$\therefore$ $f(x)$有界\\
设$M=\displaystyle\max_{x\in[0,\frac{1}{2}]}f(x)=f(c)$,$c\in[0,\frac{1}{2}]$\\
$M=f(c)=f(0)+f^{\prime}(\xi)(c-0)=f^{\prime}(\xi)c\leq f(\xi)c\leq\frac{1}{2}f(\xi)\leq\frac{1}{2}M$,$\xi \in(0,c)$\\
$\therefore$ $M=0$\\
又$\because$ $f(x)\geq0$\\
$\therefore$ $x\in [0,\frac{1}{2}]$,$f(x)\equiv 0$\\
同理$x\in[\frac{n-1}{2},\frac{n}{2}]$,有$f(x)\equiv 0$\\
$\therefore$在$[0,+\infty)$上$f(x)\equiv 0$
\end{solution}

    
\begin{problem}[本题10分]
    已知曲线$y=f(x)$在$x=0$处的切线方程为$y=10x$,令
    $g(x)=\left\{
        \begin{matrix}
            {{\frac{1}{x}\int_{0}^{1}f(x^{2}t)dt,\quad x\neq0;}}\\
            {0,\quad x=0.}\\
        \end{matrix}
        \right.$ 求$g^{\prime}(0)$.
\end{problem}
\begin{solution}[本题10分]\\
    已知$y=f(x)$在$x=0$处的切线方程为$y=10x$\\
    $\therefore$ $f(0)=0,f^{\prime}(0)=10$\\
    $g^{\prime}(0)=\displaystyle\lim_{x\to0}\frac{g(x)-g(0)}{x-0}=\displaystyle\lim_{x\to0}\frac{\int_{0}^{1}f(x^{2}t)\ d t}{x^{2}}$\\
    令$u=x^{2}t$\\
    $\therefore$ $\int_{0}^{1}f(x^{2}t)\ d t=\frac{1}{x^{2}}\int_{0}^{x^2}f(u)\ d u$\\
    $\therefore$
    $g^{\prime}(0)=\displaystyle\lim_{x\to0}\frac{\int_{0}^{x^2}f(u)\ d u}{x^{4}}=\displaystyle\lim_{x\to0}\frac{2xf(x^{2})}{4x^{3}}=\displaystyle\lim_{x\to0}\frac{f(x^{2})}{2x^{2}}\\
    =\displaystyle\lim_{x\to0}\frac{2x-f(x^{2})}{4x}=\frac{1}{2}\displaystyle\lim_{x\to0}f^{\prime}(x^{2})=\frac{1}{2}f^{\prime}(0)=5$\\

\end{solution}
    

\begin{problem}[本题15分]
    设函数$f(x)=\ln x+{\frac{1}{x}}\quad(x>0)$,\\
1)求$f(x)$的最小值;\\
2)设数列$\{x_{n}\}$满足$\ln x_{n}+\frac{1}{x_{n+1}}<1$,证明:$\displaystyle\lim_{n\to\infty}x_n$存在,并求此极限.
\end{problem}
\begin{solution}[本题15分]\\
   证明:\\
   1)$f^{\prime}(x)=\frac{1}{x}-\frac{1}{x^{2}}=\frac{x-1}{x^{2}}$\\
   令$f^{\prime}(x)=0$,则$x=1$\\
   当$x\in(0,1)$,$f^{\prime}(x)<0$,$f(x)$单减\\
   当$x\in(1,\infty)$,$f^{\prime}(x)>0$,$f(x)$单增\\
   $\therefore$ $f(x)$的最小值在极小值处取到，即$f(x)_{min}=f(1)=1$\\
   2)$f(x)=\ln x+\frac{1}{x}\geq1$\\
   $\therefore$$\ln x+\frac{1}{x_{n}}\geq1$\\
   又$\because\ln x_{n}+\frac{1}{x_{n+1}}<1$\\
   $\therefore\frac{1}{x_{n}}>\frac{1}{x_{n+1}}$\\
   $\therefore x_n<x_{n+1}$\\
   $\therefore$${x_{n}}$单增\\
   又$\because\ln x_{n}+\frac{1}{x_{n+1}}<1$\\
   $\therefore x_n<e$\\
   $\therefore$由单调有界定理知$\displaystyle\lim_{n\to\infty}x_{n}$存在，设为$a$\\
   $\therefore\displaystyle\lim_{n\to\infty}\ln(x_{n})+\frac{1}{x_n}=\ln a+\frac{1}{a}\leq1$\\
   又$\displaystyle\lim_{n\to\infty}\ln x_{n}+\frac{1}{x_{n}}=\ln a+\frac{1}{a}\geq1$\\
   $\therefore\ln a+\frac{1}{a}=1$\\
   $\therefore a=1$\\
   $\displaystyle\therefore\lim_{n\to\infty}x_{n}=1$
\end{solution}
    

\begin{problem}[本题15分]
    设函数$f(x)$在$[2,5]$上连续可微,且$f(2)=0$,证明:$\int_{2}^{5}(f^{\prime}(x))^{2}\ d x\geq\frac{M^{2}}{3}$,其中$M=\displaystyle\sup_{x\in[2,5]}f(x)$.
\end{problem}
\begin{solution}[本题15分]\\
    证明:\\
    由柯西——施瓦茨不等式知：\\
    $\int_{2}^{x}(f^{\prime}(x)^{2})\ d x\cdot\int_{2}^{x}1^{2}\ d x\geq (\int_{2}^{x}f^{\prime}(x)\ d x)^{2}=f(x)^{2}$\\
    $\therefore$对$\forall x\in[2,5]$,\\
    $f(x)^{2}\leq\int_{2}^{x}(f^{\prime}(x))^{2}\ d x\cdot\int_{2}^{x}1^{2}\ d x\leq\int_{2}^{5}(f^{\prime}(x))^{2}\ d x-\int_{2}^{5}1^{2}\ d x=3\int_{2}^{5}(f^{\prime}(x))^{2}\ d x$\\
    $\because f(x)$在$[2,5]$上连续\\
    $\therefore$在$[2,5]$上有最大值$|F(x)|\leq\sqrt{3}[\int_{2}^{5}(f^{\prime}(x))^{2}\ d t]^{\frac{1}{2}}$\\
    即$\sqrt{3}[\int_{2}^{5}(f^{\prime}(x))^{2}\ d t]^{\frac{1}{2}}$为$f(x)$的一个上界\\
    $\therefore \displaystyle M=\sup_{x\in[2,5]}f(x)\leq\sqrt{3}\left[\int_{2}^{5}(f^{\prime}(x))^{2}\ d x\right]^{\frac{1}{2}}$\\
    $\therefore M^{2}\leq3\int_{2}^{5}(f^{\prime}(x))^2\ d x$\\
    $\therefore\int_{2}^{5}f^{\prime}(x)^{2}\ d x=\frac{M^{2}}{3}$

\end{solution}

\begin{problem}[本题15分]
    设函数$f(x)$在$[0,1]$上连续,在$(0,1)$可导,且$f(0)=f(1)=0$,$f(\frac{1}{2})=1$.证明：\\
    1)存在$\eta\in\left({\frac{1}{2}},1\right)$,使得$f(\eta)=\eta $;\\
    2)对任意实数$\lambda$,一定存在$\xi \in (0,\eta)$,使得$f^{\prime}(\xi)-\lambda(f(\xi)-\xi)=1$.
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    1)设$F(x)=f(x)-x$\\
    $\therefore$$F(X)$在$[0,1]$连续,即在$(\frac{1}{2},1)$连续\\
    又$\because F(1)=f(1)-1=-1,F(\frac{1}{2})=f(\frac{1}{2})-\frac{1}{2}=\frac{1}{2}$\\
    $\therefore \eta\in(\frac{1}{2},1)$,使得$F(\eta)=f(\eta)-\eta=0$\\
    即$f(\eta)=\eta$\\
    2)构造$h(x)=e^{-\lambda x}[f(x)-x]$\\
    $h(x)$在$[0,1]$连续,$(0,1)$可导,$h(0)=h(\eta)=0$\\
    $\therefore$由罗尔中值定理$\exists \xi (0,\eta),h^{\prime}(\xi)=0$\\
    $h^{\prime}(x)=-\lambda e^{-\lambda \xi}\left[f(\xi)-\xi\right]+e^{-\lambda x}\left[f^{\prime}(x)-1\right]$\\
    $\therefore h^{\prime}(\xi)=-\lambda e^{-\lambda\xi}[f(\xi)-\xi]+e^{-\lambda\xi}[f^{\prime}(\xi)-1]=0$\\
    $\therefore f^{\prime}(\xi)-\lambda(f(\xi)-\xi)=1$
\end{solution}


\begin{problem}[本题15分]
    设$f(x)$为$(0,+\infty)$上连续减函数,$f(x)>0$,又设$a_{n}=\displaystyle\sum_{k=1}^{n}f\left(k\right)-\int_{1}^{n}f\left(x\right)\ d x$,证明${a_{n}}$为收敛数列.
\end{problem}
\begin{solution}[本题15分]
   证明:\\
   $a_{n+1}-a_{n}=\displaystyle\sum_{k=1}^{n+1}f(k)-\int_{1}^{n+1}f(x)\ d x-\displaystyle\sum_{k=1}^{n}f(k)+\int_{1}^{n}f(x)\ d x=f(n+1)-\int_{n}^{n+1}f(x)\ d x$\\
   设$F^{\prime}(x)=f(x)$\\
   则由拉格朗日中值定理$\frac{F(n+1)-F(n)}{n+1-n}=f(\xi),\xi \in (n,n+1)$\\
   $\therefore\int_{n}^{n+1}f(x)\ d x=F(n+1)-F(n)=f(\xi)$\\
   $\therefore a_{n+1}-a_{n}=f(n+1)-f(\xi)$\\
   $\because$$f(x)$单减\\
   $\therefore f(n+1)-f(\xi)<0\therefore a_{n+1}-a_n<0$\\
   $\therefore a_{n+1}<a_n<0$\\
   $\therefore{a_{n}}$单减\\
   $
   \begin{aligned}
    &a_{n}=\displaystyle\sum_{k=1}^{n}f(k)-\int_{1}^{n}f(x)\ d x \\
    &=\displaystyle\sum_{k=1}^{n}f(k)-[\int_{1}^{2}f(x)\ d x+\int_{2}^{3}f(x)\ d x+\cdots+\int_{n-1}^{n}f(x)\ d x] \\
    &=\displaystyle\sum_{k=1}^{n}f(k)-\sum_{k=1}^{n-1}\int_{k}^{k+1}f(x)\ d x \\
    &=\displaystyle\sum_{k=1}^{n-1}\left[f(k)-\int_{k}^{k+1}f(x)\ d x\right]+f(n)  \\
    &=\displaystyle\sum_{k=1}^{n-1}[f(k)-f(\eta)]+f(n)
    \end{aligned},\eta \in(k,k+1)
    $\\
    $\because f(x)$单减\\
    $\therefore f(k)-f(\eta)>0$\\
    $\therefore a_{n}=\displaystyle\sum_{k=1}^{n-1}[f(k)-f(\eta)]+f(n)>0$\\
    $\therefore$有下界\\
    $\therefore$由单调有界定理知${a_n}$极限存在,即${a_{n}}$为收敛数列
\end{solution}


\begin{problem}[本题15分]
    设求幂函数$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+1}(x-2)^{n+1}$的收敛域及和函数.
\end{problem}
\begin{solution}[本题15分]\\
    $\displaystyle \rho=\lim_{a\to \infty}\frac{a_{n+1}}{a_n}=1$\\
    $\therefore R=\frac{1}{\rho}=1$\\
    当$x=1$时，$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+1}(-1)^{n+1}$发散\\
    当$x=-1$时，$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+1}(-3)^{n+1}$发散\\
    $\therefore$收敛域为 $x-2 \in (-1,1)$,即$x \in (1,3)$\\
    令$f(x)=\displaystyle\sum_{n=1}^{n}\frac{n}{n+1}\left(x-2\right)^{n+1}$\\
    $f^{\prime}(x)=\displaystyle\sum_{n=1}^{\infty}n(x-2)^{n}=\displaystyle\sum_{n=1}^{\infty}n(x-2)^{n-1}\cdot(x-2)$\\
    令$g(x)=\displaystyle\sum_{n=1}^{n}\frac{n}{n+1}\left(x-2\right)^{n-1}$\\
    $
    \begin{gathered}
        \int_{0}^{x}g(t)\ d t=\int_{0}^{x}\displaystyle\sum_{n=1}^{\infty}n(t-2)^{n-1}\ d t=\displaystyle\sum_{n=1}^{\infty}\int_{0}^{x}n(t-2)^{n-1}\ d t \\
        =\displaystyle\sum_{n=1}^{\infty}(x-2)^{n}=\frac{x-2}{1-(x-2)}=\frac{x-2}{3-x} 
    \end{gathered}
    $\\
    $\therefore g(x)=(\frac{x-2}{3-x})^{\prime}=\frac{1}{(3-x)^{2}}$\\
    $\therefore f^{\prime}(x)=\frac{x-2}{(3-x)^{2}}$\\
    $
    \begin{aligned}
        &f(x)\\
        &=\int_{0}^{x}\frac{t-2}{(3-t)^{2}}\ d t \\
        &=-\int_{0}^{x}\frac{(3-t)-1}{(3-t)^{2}}\ d t\\
        &=-\int_{0}^{x}\frac{1}{3-t}\ d t+\int_{0}^{x}\frac{1}{(3-t)^{2}}\ d t \\
        &=\int_{0}^{x}\frac{1}{3-t}\ d (3-t)-\int_{0}^{x}\frac{1}{(3-t)^{2}}\ d (3-t) \\
        &=ln|3-t| \bigg|_{0}^{x}+\frac{1}{3-t} \bigg|_{0}^{x} \\
        &=\ln(3-x)-\ln3+\frac{1}{3-x}-\frac{1}{3}
    \end{aligned}$

\end{solution}


\begin{problem}[本题15分]
   设$W=W(x,y)$二阶可微,在$u=x^{2}-y^{2}$,$\nu=2xy$下，
    证明:$\frac{\partial^{2}W}{\partial x^{2}}+\frac{\partial^{2}W}{\partial y^{2}}=4(x^{2}+y^{2})(\frac{\partial^{2}W}{\partial u^{2}}+\frac{\partial^{2}W}{\partial\nu^{2}})$.
\end{problem}
\begin{solution}[本题15分]\\
  证明:\\
  $
  \begin{aligned}
    \frac{\partial W}{\partial x}
    &=\frac{\partial W}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial W}{\partial V}\cdot\frac{\partial V}{\partial x}\\
    &=\frac{\partial W}{\partial u}\cdot2x+\frac{\partial W}{\partial V}\cdot2y
\end{aligned}
$\\
$
\begin{gathered}
    \frac{\partial^{2}W}{\partial x^{2}}
    =[\frac{\partial^{2}W}{\partial u^{2}}\cdot 2x+\frac{\partial^{2}W}{\partial u\partial V}\cdot2y]2x+\frac{\partial W}{\partial u}\cdot2+ [\frac{\partial^{2}w}{\partial V\partial u}\cdot 2x+\frac{\partial w}{\partial V^{2}}\cdot 2y]2y+0 \\
    =\frac{\partial^{2}w}{\partial u^{2}}\cdot4x^{2}+\frac{\partial^{2}W}{\partial V^{2}}\cdot4y^{2}+8xy\frac{\partial^{2}w}{\partial u\partial V}+2\frac{\partial W}{\partial u} 
\end{gathered}
$\\
$\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial W}{\partial V}\cdot\frac{\partial V}{\partial y}=\frac{\partial W}{\partial u}\cdot(2y)+\frac{\partial W}{\partial V}\cdot2x$\\
$
\begin{gathered}
    \\
   \frac{\partial^{2}W}{\partial y^{2}}=\left[\frac{\partial W}{\partial u^{2}}\cdot-2y+\frac{\partial^{2}W}{\partial u\partial V}\cdot2x\right]-2y+\frac{\partial W}{\partial u}\cdot(-2)+ [\frac{\partial^{2}W}{\partial V\partial u}\cdot(-2y)+\frac{\partial^{2}W}{\partial V^{2}}\cdot2x]2x+0 \\
   =\frac{\partial^{2}W}{\partial u^{2}}\cdot4y^{2}+\frac{\partial^{2}W}{\partial V^{2}}\cdot4x^{2}-8xy\frac{\partial^{2}W}{\partial u\partial V}-2\frac{\partial W}{\partial u} 
\end{gathered}$\\
$\therefore\frac{\partial^{2}W}{\partial x^{2}}+\frac{\partial^{2}W}{\partial y^{2}}=4(x^{2}+y^{2})\left(\frac{\partial^{2}W}{\partial u^{2}}+\frac{\partial^{2}W}{\partial V^{2}}\right)$成立
\end{solution}
\end{document}